# (a) Derive the expression for the magnetic energy stored in an inductor when a current I develops in it. Hence, obtain the expression for the magnetic energy density.(b) A square loop of sides 5 cm. carrying a current of 0.2A. in the clockwise direction is placed at a distance of 10 cm from an infinitly long wire carrying current of 1A as shown. Calculate (i) the resultant magnetic force, and (ii) the torque, if any, acting on the loop.

(a) When the external souce supplies current to the inductor , emf is induced in it due to self induction. So, the external supply has to do work to establish current . The amount of work done is:

$d\omega = \left | \varepsilon \right |Idt$

such that, $\varepsilon = L\frac{dI}{dt}$

Therefore,

$d\omega = LIdt$

$\omega = \frac{1}{2}LI^{2}$

The energy density is equal to energy/volume.

Hence, $U= \frac{\frac{1}{2}LI^{2}}{volume}$

Consider the inductor in the form of a solinoid

then $L=\mu_0n^2Al$

Therefore

$U= \frac{\frac{1}{2}\mu_0n^2AlI^{2}}{Al}=\frac{\frac{1}{2}(\mu_0nI)^2}{\mu_0}\\\\=\frac{1}{2}\frac{B^2}{\mu_0}$

(b) We have given,

$I_{1}=1A$

$I_{2}=0.2A$

and

$r_{1}=10cm$

$r_{2}=15cm$

Now, the force of attraction experienced by length (sp) of the loop is (per unit length).

$f_{1}=\frac{\mu _{0}I_{1}I_{2}}{4\pi r_{1}}$

$f_{1}= \frac{2\times 10^{-7}\times 1\times 0.2}{10\times 10^{-2}} =4\times 10^{-7}Nm^{-1}$

the force is attractive

Now, $f_{2}=\frac{2\mu _{0}I_{1}I_{2}}{4\pi r_{2}}$

$f_{2}= \frac{2\times 10^{-7}\times 1\times 0.2}{15\times 10^{-2}} =2.6\times 10^{-7}Nm^{-1}$

the force is repulsive.

So, the net force experienced by the loop is (per unit length).

$f = \left (f_{1}-f_{2} \right )$

The total force experienced by the loop is given as:

$F = \left (f_{1}-f_{2} \right )l = \left ( 1.4\times 10^_-7 \right )\times 5\times 10^{-2}=7\times 10^{-7}N$

the resultant magnetic force is attractive in nature

The lines of action of forces coincide, that's why the torque is zero.

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